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Math Help - a better factorization?

  1. #1
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    a better factorization?

    Hey,
    I need to factor this to apply partial fractions method to a problem I'm working on. I came up with a solution, but I've always been bad at factorization and I was wondering if someone could come up with a better one.
    Here is the polynomial:
    <br />
-8x^4 + 4x^3 + 6x^2 - 5x + 1<br />
    here is the decomposition I came up with:
    -8x^4 + 4x^3 + 6x^2 - 5x + 1
    = (-8x^4 + 4x^3) + (6x^2 - 5x + 1)
    = x^3(x - 1/2) + (x - 1/3)(x - 1/2)
    = (x - 1/2)(x^3 + x - 1/3)


    Thanks!
    Last edited by epsilon; March 10th 2009 at 10:50 PM.
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  2. #2
    Junior Member lanzailan's Avatar
    Joined
    Mar 2009
    Posts
    27
    Quote Originally Posted by epsilon View Post
    Hey,
    I need to factor this to apply partial fractions method to a problem I'm working on. I came up with a solution, but I've always been bad at factorization and I was wondering if someone could come up with a better one.
    Here is the polynomial:
    <br />
-8x^4 + 4x^3 + 6x^2 - 5x + 1<br />
    here is the decomposition I came up with:
    -8x^4 + 4x^3 + 6x^2 - 5x + 1
    = (-8x^4 + 4x^3) + (6x^2 - 5x + 1)
    = x^3(x - 1/2) + (x - 1/3)(x - 1/2)
    = (x - 1/2)(x^3 + x - 1/3)


    Thanks!
    i've got a better way to solve this problem..
    -8x^4 + 4x^3 + 6x^2 - 5x + 1
    -(8x^4-4x^3)+(3x-1)(2x-1)
    -4x^3(2x-1)+(3x-1)(2x-1)
    -(2x-1)[4x^3-(3x-1)]
    -(2x-1)(4x^3-3x+1)

    ok, now we have to use try and error method to find another one root..

    (4x^3-3x+1)=0
    let's plug in  x=-1
    4(-1)^3-3(-1)+1=0
    LHS=RHS
    so -1 is the another root or (x-1)
    then perform long division
    (x-1)/(4x^3-3x+1)
    ...
    ...
    -(2x-1)(x-1)(4x^2-4x+1)
    -(2x-1)(x-1)(2x-1)^2
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